\(\left\{{}\begin{matrix}\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\\\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3\left(2x+1\right)-4\left(y-2\right)}{12}=\dfrac{1}{12}\\\dfrac{3\left(x+5\right)-2\left(y+7\right)}{6}=-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3\left(2x+1\right)-4\left(y-2\right)=1\\3\left(x+5\right)-2\left(y+7\right)=-24\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x+3-4y+8=1\\3x+15-2y-14=-24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-4y=1-3-8=-10\\3x-2y=-24-1=-25\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-2y=-5\\3x-2y=-5\end{matrix}\right.\left(vôlý\right)\)
Vậy: Hệ vô nghiệm
